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Pg 79 of "Tensors, Relativity and Cosmology"

In order to construct the geodesic equations which define the curve with a stationary arc length, we may choose the arc length itself as the action integral with zero variation. Using $$s(t)=\int_{t_o} ^t\sqrt{g_{mn}\frac{dx^m}{dt}\frac{dx^n}{dt}}dt \tag{1}$$

we write $$I=\int_{S_A}^{S_B}\sqrt{g_{mn}\frac{dx^m}{ds}\frac{dx^n}{ds}}ds \tag{2}$$

where $I$ is the action integral and is equal to unity along the geodesic line C, where $$ds^2=g_{mn}dx^mdx^n \tag{3}$$

But isn't (2) just equal to 1 since $\frac{ds}{ds}=1$ and can this still be an action integral (functional) if the LHS of (2) is just a constant?

Or was it the author's intention to introduce to a scalar parameter such as $d\lambda$ but accidentally used ds instead?

(The author actually made tons of mistakes in the previous chapters, so I'm just making sure that this is not another error but I could be wrong...)

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  • $\begingroup$ Or to put it another way, under what circumstances is (3) true? $\endgroup$
    – puppetsock
    Commented Jun 14, 2019 at 14:34

2 Answers 2

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Yes, OP is right. One cannot consistently formulate a variational principle for geodesics with arc length/affine parameter as the independent curve parameter for all virtual curves. Although the square root action is indeed independent of parametrization $\lambda$, the point is that the parameter interval $[\lambda_i,\lambda_f]$ must be common for all virtual curves, so that if $\lambda$ happens to be the arc length/affine parameter for one particular curve, this will not be the case for all neighboring curves. For more details, see e.g. my Phys.SE answer here.

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    $\begingroup$ The author might (admittedly confusingly) claim that the notation $s$ in eq. (1) denotes arc length while the notation $s$ in eq. (2) denotes an arbitrary parameter. $\endgroup$
    – Qmechanic
    Commented Jun 14, 2019 at 11:49
  • $\begingroup$ And this arbitrary parameter cannot be/is not necessary an affine parameter/arc length parameter? $\endgroup$ Commented Jun 14, 2019 at 13:14
  • $\begingroup$ Not for all virtual curves. $\endgroup$
    – Qmechanic
    Commented Jun 14, 2019 at 19:08
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One can choose any parameter one likes to parameterize a curve and the standard trick is to use $s$ itself (sometimes called "proper time"). So the length of a curve is defined to be $$ L = \int_{t_0}^{t_1}\sqrt{g_{ab}\frac{dx^a}{dt}\frac{dx^b}{dt}}\,dt $$ but if we choose $t=s$ we have $$ L = \int_{s_0}^{s_1} ds = s_1-s_0\,. $$ This is called "proper time" or sometimes "unit speed". This does not help compute the length of the curve however, since we still need to evaluate $s_1-s_0$.

The geodesic equations arise from minimizing an action. There are two equivalent choices, one can minimize the length $L$ or one can also minimize the energy functional $$ I = \int_{t_0}^{t_1}g_{ab}\frac{dx^a}{dt}\frac{dx^b}{dt}\,dt $$ The actions $L$ and $I$ are different but their variations both lead to the geodesic equations.

Not sure I answered exactly the question, the text your using sounds a little imprecise (maybe try Sean Carroll's book Spacetime and Geometry), but I hope it helps.

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